3.84 \(\int \frac {\csc ^5(c+d x)}{a+b \sin ^2(c+d x)} \, dx\)

Optimal. Leaf size=125 \[ \frac {b^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \cos (c+d x)}{\sqrt {a+b}}\right )}{a^3 d \sqrt {a+b}}-\frac {(3 a-4 b) \cot (c+d x) \csc (c+d x)}{8 a^2 d}-\frac {\left (3 a^2-4 a b+8 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{8 a^3 d}-\frac {\cot (c+d x) \csc ^3(c+d x)}{4 a d} \]

[Out]

-1/8*(3*a^2-4*a*b+8*b^2)*arctanh(cos(d*x+c))/a^3/d-1/8*(3*a-4*b)*cot(d*x+c)*csc(d*x+c)/a^2/d-1/4*cot(d*x+c)*cs
c(d*x+c)^3/a/d+b^(5/2)*arctanh(cos(d*x+c)*b^(1/2)/(a+b)^(1/2))/a^3/d/(a+b)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.19, antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3186, 414, 527, 522, 206, 208} \[ -\frac {\left (3 a^2-4 a b+8 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{8 a^3 d}+\frac {b^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \cos (c+d x)}{\sqrt {a+b}}\right )}{a^3 d \sqrt {a+b}}-\frac {(3 a-4 b) \cot (c+d x) \csc (c+d x)}{8 a^2 d}-\frac {\cot (c+d x) \csc ^3(c+d x)}{4 a d} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^5/(a + b*Sin[c + d*x]^2),x]

[Out]

-((3*a^2 - 4*a*b + 8*b^2)*ArcTanh[Cos[c + d*x]])/(8*a^3*d) + (b^(5/2)*ArcTanh[(Sqrt[b]*Cos[c + d*x])/Sqrt[a +
b]])/(a^3*Sqrt[a + b]*d) - ((3*a - 4*b)*Cot[c + d*x]*Csc[c + d*x])/(8*a^2*d) - (Cot[c + d*x]*Csc[c + d*x]^3)/(
4*a*d)

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 414

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*x*(a + b*x^n)^(p + 1)*(
c + d*x^n)^(q + 1))/(a*n*(p + 1)*(b*c - a*d)), x] + Dist[1/(a*n*(p + 1)*(b*c - a*d)), Int[(a + b*x^n)^(p + 1)*
(c + d*x^n)^q*Simp[b*c + n*(p + 1)*(b*c - a*d) + d*b*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d,
n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  !( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomial
Q[a, b, c, d, n, p, q, x]

Rule 522

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]

Rule 527

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> -Simp[
((b*e - a*f)*x*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*n*(b*c - a*d)*(p + 1)), x] + Dist[1/(a*n*(b*c - a*d
)*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*f)
*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, q}, x] && LtQ[p, -1]

Rule 3186

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Cos[e + f*x], x]}, -Dist[ff/f, Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b - b*ff^2*x^2)^p, x], x, Cos
[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rubi steps

\begin {align*} \int \frac {\csc ^5(c+d x)}{a+b \sin ^2(c+d x)} \, dx &=-\frac {\operatorname {Subst}\left (\int \frac {1}{\left (1-x^2\right )^3 \left (a+b-b x^2\right )} \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac {\cot (c+d x) \csc ^3(c+d x)}{4 a d}-\frac {\operatorname {Subst}\left (\int \frac {3 a-b-3 b x^2}{\left (1-x^2\right )^2 \left (a+b-b x^2\right )} \, dx,x,\cos (c+d x)\right )}{4 a d}\\ &=-\frac {(3 a-4 b) \cot (c+d x) \csc (c+d x)}{8 a^2 d}-\frac {\cot (c+d x) \csc ^3(c+d x)}{4 a d}-\frac {\operatorname {Subst}\left (\int \frac {3 a^2-a b+4 b^2-(3 a-4 b) b x^2}{\left (1-x^2\right ) \left (a+b-b x^2\right )} \, dx,x,\cos (c+d x)\right )}{8 a^2 d}\\ &=-\frac {(3 a-4 b) \cot (c+d x) \csc (c+d x)}{8 a^2 d}-\frac {\cot (c+d x) \csc ^3(c+d x)}{4 a d}+\frac {b^3 \operatorname {Subst}\left (\int \frac {1}{a+b-b x^2} \, dx,x,\cos (c+d x)\right )}{a^3 d}-\frac {\left (3 a^2-4 a b+8 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\cos (c+d x)\right )}{8 a^3 d}\\ &=-\frac {\left (3 a^2-4 a b+8 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{8 a^3 d}+\frac {b^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \cos (c+d x)}{\sqrt {a+b}}\right )}{a^3 \sqrt {a+b} d}-\frac {(3 a-4 b) \cot (c+d x) \csc (c+d x)}{8 a^2 d}-\frac {\cot (c+d x) \csc ^3(c+d x)}{4 a d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 6.30, size = 657, normalized size = 5.26 \[ \frac {b^{5/2} \csc ^2(c+d x) (-2 a+b \cos (2 (c+d x))-b) \tan ^{-1}\left (\frac {\sec \left (\frac {1}{2} (c+d x)\right ) \left (\sqrt {b} \cos \left (\frac {1}{2} (c+d x)\right )-i \sqrt {a} \sin \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {-a-b}}\right )}{2 a^3 d \sqrt {-a-b} \left (a \csc ^2(c+d x)+b\right )}+\frac {b^{5/2} \csc ^2(c+d x) (-2 a+b \cos (2 (c+d x))-b) \tan ^{-1}\left (\frac {\sec \left (\frac {1}{2} (c+d x)\right ) \left (\sqrt {b} \cos \left (\frac {1}{2} (c+d x)\right )+i \sqrt {a} \sin \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {-a-b}}\right )}{2 a^3 d \sqrt {-a-b} \left (a \csc ^2(c+d x)+b\right )}+\frac {(3 a-4 b) \csc ^2(c+d x) \csc ^2\left (\frac {1}{2} (c+d x)\right ) (-2 a+b \cos (2 (c+d x))-b)}{64 a^2 d \left (a \csc ^2(c+d x)+b\right )}+\frac {(4 b-3 a) \csc ^2(c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right ) (-2 a+b \cos (2 (c+d x))-b)}{64 a^2 d \left (a \csc ^2(c+d x)+b\right )}+\frac {\left (3 a^2-4 a b+8 b^2\right ) \csc ^2(c+d x) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) (-2 a+b \cos (2 (c+d x))-b)}{16 a^3 d \left (a \csc ^2(c+d x)+b\right )}+\frac {\left (-3 a^2+4 a b-8 b^2\right ) \csc ^2(c+d x) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) (-2 a+b \cos (2 (c+d x))-b)}{16 a^3 d \left (a \csc ^2(c+d x)+b\right )}+\frac {\csc ^2(c+d x) \csc ^4\left (\frac {1}{2} (c+d x)\right ) (-2 a+b \cos (2 (c+d x))-b)}{128 a d \left (a \csc ^2(c+d x)+b\right )}-\frac {\csc ^2(c+d x) \sec ^4\left (\frac {1}{2} (c+d x)\right ) (-2 a+b \cos (2 (c+d x))-b)}{128 a d \left (a \csc ^2(c+d x)+b\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]^5/(a + b*Sin[c + d*x]^2),x]

[Out]

(b^(5/2)*ArcTan[(Sec[(c + d*x)/2]*(Sqrt[b]*Cos[(c + d*x)/2] - I*Sqrt[a]*Sin[(c + d*x)/2]))/Sqrt[-a - b]]*(-2*a
- b + b*Cos[2*(c + d*x)])*Csc[c + d*x]^2)/(2*a^3*Sqrt[-a - b]*d*(b + a*Csc[c + d*x]^2)) + (b^(5/2)*ArcTan[(Se
c[(c + d*x)/2]*(Sqrt[b]*Cos[(c + d*x)/2] + I*Sqrt[a]*Sin[(c + d*x)/2]))/Sqrt[-a - b]]*(-2*a - b + b*Cos[2*(c +
d*x)])*Csc[c + d*x]^2)/(2*a^3*Sqrt[-a - b]*d*(b + a*Csc[c + d*x]^2)) + ((3*a - 4*b)*(-2*a - b + b*Cos[2*(c +
d*x)])*Csc[(c + d*x)/2]^2*Csc[c + d*x]^2)/(64*a^2*d*(b + a*Csc[c + d*x]^2)) + ((-2*a - b + b*Cos[2*(c + d*x)])
*Csc[(c + d*x)/2]^4*Csc[c + d*x]^2)/(128*a*d*(b + a*Csc[c + d*x]^2)) + ((3*a^2 - 4*a*b + 8*b^2)*(-2*a - b + b*
Cos[2*(c + d*x)])*Csc[c + d*x]^2*Log[Cos[(c + d*x)/2]])/(16*a^3*d*(b + a*Csc[c + d*x]^2)) + ((-3*a^2 + 4*a*b -
8*b^2)*(-2*a - b + b*Cos[2*(c + d*x)])*Csc[c + d*x]^2*Log[Sin[(c + d*x)/2]])/(16*a^3*d*(b + a*Csc[c + d*x]^2)
) + ((-3*a + 4*b)*(-2*a - b + b*Cos[2*(c + d*x)])*Csc[c + d*x]^2*Sec[(c + d*x)/2]^2)/(64*a^2*d*(b + a*Csc[c +
d*x]^2)) - ((-2*a - b + b*Cos[2*(c + d*x)])*Csc[c + d*x]^2*Sec[(c + d*x)/2]^4)/(128*a*d*(b + a*Csc[c + d*x]^2)
)

________________________________________________________________________________________

fricas [B]  time = 0.51, size = 612, normalized size = 4.90 \[ \left [\frac {2 \, {\left (3 \, a^{2} - 4 \, a b\right )} \cos \left (d x + c\right )^{3} + 8 \, {\left (b^{2} \cos \left (d x + c\right )^{4} - 2 \, b^{2} \cos \left (d x + c\right )^{2} + b^{2}\right )} \sqrt {\frac {b}{a + b}} \log \left (\frac {b \cos \left (d x + c\right )^{2} + 2 \, {\left (a + b\right )} \sqrt {\frac {b}{a + b}} \cos \left (d x + c\right ) + a + b}{b \cos \left (d x + c\right )^{2} - a - b}\right ) - 2 \, {\left (5 \, a^{2} - 4 \, a b\right )} \cos \left (d x + c\right ) - {\left ({\left (3 \, a^{2} - 4 \, a b + 8 \, b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (3 \, a^{2} - 4 \, a b + 8 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \, a^{2} - 4 \, a b + 8 \, b^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + {\left ({\left (3 \, a^{2} - 4 \, a b + 8 \, b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (3 \, a^{2} - 4 \, a b + 8 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \, a^{2} - 4 \, a b + 8 \, b^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{16 \, {\left (a^{3} d \cos \left (d x + c\right )^{4} - 2 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d\right )}}, \frac {2 \, {\left (3 \, a^{2} - 4 \, a b\right )} \cos \left (d x + c\right )^{3} - 16 \, {\left (b^{2} \cos \left (d x + c\right )^{4} - 2 \, b^{2} \cos \left (d x + c\right )^{2} + b^{2}\right )} \sqrt {-\frac {b}{a + b}} \arctan \left (\sqrt {-\frac {b}{a + b}} \cos \left (d x + c\right )\right ) - 2 \, {\left (5 \, a^{2} - 4 \, a b\right )} \cos \left (d x + c\right ) - {\left ({\left (3 \, a^{2} - 4 \, a b + 8 \, b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (3 \, a^{2} - 4 \, a b + 8 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \, a^{2} - 4 \, a b + 8 \, b^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + {\left ({\left (3 \, a^{2} - 4 \, a b + 8 \, b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (3 \, a^{2} - 4 \, a b + 8 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \, a^{2} - 4 \, a b + 8 \, b^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{16 \, {\left (a^{3} d \cos \left (d x + c\right )^{4} - 2 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^5/(a+b*sin(d*x+c)^2),x, algorithm="fricas")

[Out]

[1/16*(2*(3*a^2 - 4*a*b)*cos(d*x + c)^3 + 8*(b^2*cos(d*x + c)^4 - 2*b^2*cos(d*x + c)^2 + b^2)*sqrt(b/(a + b))*
log((b*cos(d*x + c)^2 + 2*(a + b)*sqrt(b/(a + b))*cos(d*x + c) + a + b)/(b*cos(d*x + c)^2 - a - b)) - 2*(5*a^2
- 4*a*b)*cos(d*x + c) - ((3*a^2 - 4*a*b + 8*b^2)*cos(d*x + c)^4 - 2*(3*a^2 - 4*a*b + 8*b^2)*cos(d*x + c)^2 +
3*a^2 - 4*a*b + 8*b^2)*log(1/2*cos(d*x + c) + 1/2) + ((3*a^2 - 4*a*b + 8*b^2)*cos(d*x + c)^4 - 2*(3*a^2 - 4*a*
b + 8*b^2)*cos(d*x + c)^2 + 3*a^2 - 4*a*b + 8*b^2)*log(-1/2*cos(d*x + c) + 1/2))/(a^3*d*cos(d*x + c)^4 - 2*a^3
*d*cos(d*x + c)^2 + a^3*d), 1/16*(2*(3*a^2 - 4*a*b)*cos(d*x + c)^3 - 16*(b^2*cos(d*x + c)^4 - 2*b^2*cos(d*x +
c)^2 + b^2)*sqrt(-b/(a + b))*arctan(sqrt(-b/(a + b))*cos(d*x + c)) - 2*(5*a^2 - 4*a*b)*cos(d*x + c) - ((3*a^2
- 4*a*b + 8*b^2)*cos(d*x + c)^4 - 2*(3*a^2 - 4*a*b + 8*b^2)*cos(d*x + c)^2 + 3*a^2 - 4*a*b + 8*b^2)*log(1/2*co
s(d*x + c) + 1/2) + ((3*a^2 - 4*a*b + 8*b^2)*cos(d*x + c)^4 - 2*(3*a^2 - 4*a*b + 8*b^2)*cos(d*x + c)^2 + 3*a^2
- 4*a*b + 8*b^2)*log(-1/2*cos(d*x + c) + 1/2))/(a^3*d*cos(d*x + c)^4 - 2*a^3*d*cos(d*x + c)^2 + a^3*d)]

________________________________________________________________________________________

giac [B]  time = 0.25, size = 334, normalized size = 2.67 \[ -\frac {\frac {64 \, b^{3} \arctan \left (\frac {b \cos \left (d x + c\right ) + a + b}{\sqrt {-a b - b^{2}} \cos \left (d x + c\right ) + \sqrt {-a b - b^{2}}}\right )}{\sqrt {-a b - b^{2}} a^{3}} + \frac {\frac {8 \, a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {8 \, b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {a {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}}{a^{2}} - \frac {4 \, {\left (3 \, a^{2} - 4 \, a b + 8 \, b^{2}\right )} \log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right )}{a^{3}} + \frac {{\left (a^{2} - \frac {8 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {8 \, a b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {18 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {24 \, a b {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {48 \, b^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}^{2}}{a^{3} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}}{64 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^5/(a+b*sin(d*x+c)^2),x, algorithm="giac")

[Out]

-1/64*(64*b^3*arctan((b*cos(d*x + c) + a + b)/(sqrt(-a*b - b^2)*cos(d*x + c) + sqrt(-a*b - b^2)))/(sqrt(-a*b -
b^2)*a^3) + (8*a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 8*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - a*(cos(d
*x + c) - 1)^2/(cos(d*x + c) + 1)^2)/a^2 - 4*(3*a^2 - 4*a*b + 8*b^2)*log(abs(-cos(d*x + c) + 1)/abs(cos(d*x +
c) + 1))/a^3 + (a^2 - 8*a^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 8*a*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1
) + 18*a^2*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 - 24*a*b*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 48*b
^2*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1)^2/(a^3*(cos(d*x + c) - 1)^2))/d

________________________________________________________________________________________

maple [B]  time = 0.52, size = 255, normalized size = 2.04 \[ -\frac {1}{16 a d \left (\cos \left (d x +c \right )-1\right )^{2}}+\frac {3}{16 a d \left (\cos \left (d x +c \right )-1\right )}-\frac {b}{4 d \,a^{2} \left (\cos \left (d x +c \right )-1\right )}+\frac {3 \ln \left (\cos \left (d x +c \right )-1\right )}{16 a d}-\frac {\ln \left (\cos \left (d x +c \right )-1\right ) b}{4 d \,a^{2}}+\frac {\ln \left (\cos \left (d x +c \right )-1\right ) b^{2}}{2 d \,a^{3}}+\frac {b^{3} \arctanh \left (\frac {\cos \left (d x +c \right ) b}{\sqrt {\left (a +b \right ) b}}\right )}{d \,a^{3} \sqrt {\left (a +b \right ) b}}+\frac {1}{16 a d \left (1+\cos \left (d x +c \right )\right )^{2}}+\frac {3}{16 a d \left (1+\cos \left (d x +c \right )\right )}-\frac {b}{4 d \,a^{2} \left (1+\cos \left (d x +c \right )\right )}-\frac {3 \ln \left (1+\cos \left (d x +c \right )\right )}{16 a d}+\frac {\ln \left (1+\cos \left (d x +c \right )\right ) b}{4 d \,a^{2}}-\frac {\ln \left (1+\cos \left (d x +c \right )\right ) b^{2}}{2 d \,a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^5/(a+b*sin(d*x+c)^2),x)

[Out]

-1/16/a/d/(cos(d*x+c)-1)^2+3/16/a/d/(cos(d*x+c)-1)-1/4/d/a^2/(cos(d*x+c)-1)*b+3/16/a/d*ln(cos(d*x+c)-1)-1/4/d/
a^2*ln(cos(d*x+c)-1)*b+1/2/d/a^3*ln(cos(d*x+c)-1)*b^2+1/d*b^3/a^3/((a+b)*b)^(1/2)*arctanh(cos(d*x+c)*b/((a+b)*
b)^(1/2))+1/16/a/d/(1+cos(d*x+c))^2+3/16/a/d/(1+cos(d*x+c))-1/4/d/a^2/(1+cos(d*x+c))*b-3/16/a/d*ln(1+cos(d*x+c
))+1/4/d/a^2*ln(1+cos(d*x+c))*b-1/2/d/a^3*ln(1+cos(d*x+c))*b^2

________________________________________________________________________________________

maxima [A]  time = 0.43, size = 181, normalized size = 1.45 \[ -\frac {\frac {8 \, b^{3} \log \left (\frac {b \cos \left (d x + c\right ) - \sqrt {{\left (a + b\right )} b}}{b \cos \left (d x + c\right ) + \sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {{\left (a + b\right )} b} a^{3}} - \frac {2 \, {\left ({\left (3 \, a - 4 \, b\right )} \cos \left (d x + c\right )^{3} - {\left (5 \, a - 4 \, b\right )} \cos \left (d x + c\right )\right )}}{a^{2} \cos \left (d x + c\right )^{4} - 2 \, a^{2} \cos \left (d x + c\right )^{2} + a^{2}} + \frac {{\left (3 \, a^{2} - 4 \, a b + 8 \, b^{2}\right )} \log \left (\cos \left (d x + c\right ) + 1\right )}{a^{3}} - \frac {{\left (3 \, a^{2} - 4 \, a b + 8 \, b^{2}\right )} \log \left (\cos \left (d x + c\right ) - 1\right )}{a^{3}}}{16 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^5/(a+b*sin(d*x+c)^2),x, algorithm="maxima")

[Out]

-1/16*(8*b^3*log((b*cos(d*x + c) - sqrt((a + b)*b))/(b*cos(d*x + c) + sqrt((a + b)*b)))/(sqrt((a + b)*b)*a^3)
- 2*((3*a - 4*b)*cos(d*x + c)^3 - (5*a - 4*b)*cos(d*x + c))/(a^2*cos(d*x + c)^4 - 2*a^2*cos(d*x + c)^2 + a^2)
+ (3*a^2 - 4*a*b + 8*b^2)*log(cos(d*x + c) + 1)/a^3 - (3*a^2 - 4*a*b + 8*b^2)*log(cos(d*x + c) - 1)/a^3)/d

________________________________________________________________________________________

mupad [B]  time = 13.93, size = 1105, normalized size = 8.84 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(c + d*x)^5*(a + b*sin(c + d*x)^2)),x)

[Out]

- ((cos(c + d*x)*(5*a - 4*b))/(8*a^2) - (cos(c + d*x)^3*(3*a - 4*b))/(8*a^2))/(d*(cos(c + d*x)^4 - cos(c + d*x
)^2 + sin(c + d*x)^2)) - (atanh((63*b^4*cos(c + d*x))/(64*((63*b^4)/64 - (81*a*b^3)/256 + (27*a^2*b^2)/256 - (
35*b^5)/(32*a) + (5*b^6)/(4*a^2))) - (81*b^3*cos(c + d*x))/(256*((27*a*b^2)/256 - (81*b^3)/256 + (63*b^4)/(64*
a) - (35*b^5)/(32*a^2) + (5*b^6)/(4*a^3))) - (35*b^5*cos(c + d*x))/(32*((63*a*b^4)/64 - (35*b^5)/32 - (81*a^2*
b^3)/256 + (27*a^3*b^2)/256 + (5*b^6)/(4*a))) + (5*b^6*cos(c + d*x))/(4*((5*b^6)/4 - (35*a*b^5)/32 + (63*a^2*b
^4)/64 - (81*a^3*b^3)/256 + (27*a^4*b^2)/256)) + (27*b^2*cos(c + d*x))/(256*((27*b^2)/256 - (81*b^3)/(256*a) +
(63*b^4)/(64*a^2) - (35*b^5)/(32*a^3) + (5*b^6)/(4*a^4))))*(3*a^2 - 4*a*b + 8*b^2))/(8*a^3*d) - (atan((((b^5*
(a + b))^(1/2)*((cos(c + d*x)*(128*b^7 - 64*a*b^6 + 64*a^2*b^5 - 24*a^3*b^4 + 9*a^4*b^3))/(64*a^4) + ((b^5*(a
+ b))^(1/2)*((2*a^6*b^4 - (a^7*b^3)/2 + (3*a^8*b^2)/2)/(2*a^6) - (cos(c + d*x)*(512*a^6*b^3 + 256*a^7*b^2)*(b^
5*(a + b))^(1/2))/(128*a^4*(a^3*b + a^4))))/(2*(a^3*b + a^4)))*1i)/(a^3*b + a^4) + ((b^5*(a + b))^(1/2)*((cos(
c + d*x)*(128*b^7 - 64*a*b^6 + 64*a^2*b^5 - 24*a^3*b^4 + 9*a^4*b^3))/(64*a^4) - ((b^5*(a + b))^(1/2)*((2*a^6*b
^4 - (a^7*b^3)/2 + (3*a^8*b^2)/2)/(2*a^6) + (cos(c + d*x)*(512*a^6*b^3 + 256*a^7*b^2)*(b^5*(a + b))^(1/2))/(12
8*a^4*(a^3*b + a^4))))/(2*(a^3*b + a^4)))*1i)/(a^3*b + a^4))/(((5*a*b^7)/4 - b^8 - (3*a^2*b^6)/4 + (9*a^3*b^5)
/32)/a^6 + ((b^5*(a + b))^(1/2)*((cos(c + d*x)*(128*b^7 - 64*a*b^6 + 64*a^2*b^5 - 24*a^3*b^4 + 9*a^4*b^3))/(64
*a^4) + ((b^5*(a + b))^(1/2)*((2*a^6*b^4 - (a^7*b^3)/2 + (3*a^8*b^2)/2)/(2*a^6) - (cos(c + d*x)*(512*a^6*b^3 +
256*a^7*b^2)*(b^5*(a + b))^(1/2))/(128*a^4*(a^3*b + a^4))))/(2*(a^3*b + a^4))))/(a^3*b + a^4) - ((b^5*(a + b)
)^(1/2)*((cos(c + d*x)*(128*b^7 - 64*a*b^6 + 64*a^2*b^5 - 24*a^3*b^4 + 9*a^4*b^3))/(64*a^4) - ((b^5*(a + b))^(
1/2)*((2*a^6*b^4 - (a^7*b^3)/2 + (3*a^8*b^2)/2)/(2*a^6) + (cos(c + d*x)*(512*a^6*b^3 + 256*a^7*b^2)*(b^5*(a +
b))^(1/2))/(128*a^4*(a^3*b + a^4))))/(2*(a^3*b + a^4))))/(a^3*b + a^4)))*(b^5*(a + b))^(1/2)*1i)/(d*(a^3*b + a
^4))

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)**5/(a+b*sin(d*x+c)**2),x)

[Out]

Timed out

________________________________________________________________________________________